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pentagon [3]
3 years ago
12

Myra multiplies 5 integers. The product is negative. AT LEAST, how many of the integers must be negative?

Mathematics
2 answers:
kicyunya [14]3 years ago
7 0

1, If 0 are negative the product is positive, but if 2 are negative then answer is still positive. At least 1 but could also be 1,3,5.

KonstantinChe [14]3 years ago
4 0

1, If 0 are negative the product is positive

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Use the graphs of f and g to solve Exercises 87, 88, and 89.
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Answers and explanations:

87. The domain of added functions includes the restrictions of both. So the range of the added function in this question is [-4, 3]

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What expression best estimates -18 1/4 ÷ 2 2/3<br>a. 18÷3<br>b. -18÷3<br>c. -18÷(-3)<br>d. 18÷(-3)​
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bija089 [108]

Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

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Step-by-step explanation:

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